Guides / Inequalities

Absolute value inequalities: when it’s “and”, when it’s “or”

Less-than means the answers sit between two numbers; greater-than means they sit outside them. Most lost points come from mixing those up, from splitting before the bars are alone, and from a missed flip in one of the cases.

Why it works

|u| is the distance from u to 0 on the number line. |u| < 3 says "u is less than 3 units from 0", and those numbers fill the stretch between −3 and 3: one connected piece, so it's an "and", −3 < u < 3. |u| > 3 says "more than 3 units from 0", which happens in two separate places, off to the left past −3 or off to the right past 3. A number can't be in both places at once, so it's an "or".

The distance reading also works with a shifted center. |x − 5| ≤ 2 means "x is within 2 of 5", so 3 ≤ x ≤ 7 without any algebra. That makes a good check on the algebra you do write.

A distance can never be negative. That's why a negative k is a special case: no distance is less than −2, and every distance is greater than −2.

Example 1: less-than, so "between"

Solve |2x − 1| < 7.

1. −7 < 2x − 1 < 7The bars are alone and 7 is positive. Less-than means 2x − 1 is trapped between −7 and 7.
2. −6 < 2x < 8Add 1 to all three parts.
3. −3 < x < 4Divide all three parts by 2. Positive, so no flip. In interval notation: (−3, 4).

Check one number inside and one outside. x = 0: |−1| = 1 < 7, true. x = 5: |9| = 9, not less than 7. The endpoints are excluded because at x = 4, |7| = 7, which isn't less than 7.

Example 2: greater-than, so "or"

Solve |x + 3| ≥ 5.

1. x + 3 ≤ −5 or x + 3 ≥ 5Greater-than means x + 3 is at least 5 away from 0, in either direction. Two separate inequalities.
2. x ≤ −8 or x ≥ 2Subtract 3 from each.
3. (−∞, −8] ∪ [2, ∞)Interval notation. Square brackets because ≥ includes the endpoints; ∪ is how "or" is written between intervals.

Check: x = −10 gives |−7| = 7 ≥ 5, true. x = 0 gives |3| = 3, false, and 0 is correctly left out.

Example 3: isolate first, then flip inside a case

Solve 2|3 − x| + 1 > 9.

1. 2|3 − x| > 8Subtract 1. The 2 and the 1 are outside the bars, so they have to go before you split.
2. |3 − x| > 4Divide by 2. Positive, so no flip. Now the bars are alone.
3. 3 − x < −4 or 3 − x > 4Greater-than: two outward pieces.
4. −x < −7 or −x > 1Subtract 3 from each.
5. x > 7 or x < −1Divide each by −1, so flip both signs.
6. (−∞, −1) ∪ (7, ∞)Smallest piece first. Parentheses because > excludes the endpoints.

Distance check: |3 − x| is the same as |x − 3|, the distance from x to 3. More than 4 away from 3 means below −1 or above 7, which matches.

Example 4: a negative number on the other side

Solve |4x + 1| + 6 < 2, and then |4x + 1| + 6 > 2.

1. |4x + 1| < −4Subtract 6. The absolute value would have to be less than a negative number.
2. no solutionAn absolute value is never negative, so it is never below −4. In interval notation, the empty set ∅.
3. |4x + 1| > −4The second inequality, after subtracting 6.
4. all real numbersEvery absolute value is 0 or more, so every one is greater than −4. In interval notation, (−∞, ∞).

Check the "all real numbers" answer at the point where it seems most likely to fail, where the inside is 0: x = −14 gives |0| + 6 = 6 > 2. True.

Common mistakes, and the exact line they happen on

1. Turning a less-than into an "or"

Solve |2x − 1| < 7 (Example 1).

1. 2x − 1 < 7 or 2x − 1 > −7This is the mistake. These two pieces overlap and together cover every number, so this leads to x < 4 or x > −3, which is all real numbers. But x = 10 gives |19| = 19, which isn't less than 7. The two conditions both have to hold: −7 < 2x − 1 < 7.

If your "or" answer covers the whole number line, the inequality should have been an "and". Less-than is always one piece in the middle.

2. Writing the greater-than case as one chain

Solve |x + 3| ≥ 5 (Example 2).

1. −5 ≤ x + 3 ≥ 5This is the mistake. A chain like this says x + 3 is at least −5 and at least 5, which is just x + 3 ≥ 5. It leads to x ≥ 2 and loses the whole left piece. x = −10 gives |−7| = 7 ≥ 5, so it's a solution this answer throws away.

A chain a < u < b only makes sense when it describes one piece between two numbers. Greater-than gives two pieces, so it needs two inequalities and the word "or".

3. Forgetting to flip in the negative case

Solve |3 − x| > 4 (Example 3).

1. −x < −7 or −x > 1Correct so far.
2. x < 7 or x < −1This is the mistake. The first piece divided by −1 without flipping. Together the pieces say x < 7, which includes x = 0, where |3 − 0| = 3 is not greater than 4.

Each case is its own ordinary inequality, and the flip rule applies in each one. If both of your pieces point the same way, a flip went missing; a greater-than answer should always point outward in both directions.

4. Splitting before isolating

Solve 2|3 − x| + 1 > 9 (Example 3).

1. 2(3 − x) + 1 < −9 or
   2(3 − x) + 1 > 9This is the mistake. The −9 applies the negative case to the 2 and the +1, which sit outside the bars. This leads to x > 8 or x < −1, which wrongly leaves out numbers like x = 7.5: 2|3 − 7.5| + 1 = 10, which is greater than 9.

Only what's inside the bars gets the ±. Move everything else away first, so the line just before you split reads |stuff| < k or |stuff| > k and nothing more.

5. Treating every negative k as "no solution"

1. |x − 2| + 5 > 2Start.
2. |x − 2| > −3Subtract 5. Correct.
3. no solutionThis is the mistake. That's the right answer for |x − 2| = −3 or |x − 2| < −3, but this one asks whether an absolute value can be greater than −3, and it always is. Try x = 2: |0| = 0 > −3. The answer is all real numbers.

When k is negative, don't split and don't guess. Ask whether a number that's never negative can be less than, or greater than, a negative number. Less-than: never, so no solution. Greater-than: always, so all real numbers.

Practice

1. Solve |x − 4| ≤ 6. Give the answer in interval notation.

   Show answer

   −6 ≤ x − 4 ≤ 6, so −2 ≤ x ≤ 10, which is [−2, 10]. Distance check: within 6 of 4.

2. Solve |2x + 5| > 3.

   Show answer

   2x + 5 < −3 or 2x + 5 > 3, so x < −4 or x > −1, which is (−∞, −4) ∪ (−1, ∞).

3. Solve 3|x − 1| − 2 < 10.

   Show answer

   Isolate: 3|x − 1| < 12, |x − 1| < 4. Then −4 < x − 1 < 4, so −3 < x < 5, which is (−3, 5).

4. Solve |5 − 2x| ≥ 9.

   Show answer

   5 − 2x ≤ −9 or 5 − 2x ≥ 9. Subtract 5: −2x ≤ −14 or −2x ≥ 4. Divide by −2 and flip both: x ≥ 7 or x ≤ −2, which is (−∞, −2] ∪ [7, ∞).

5. Solve |x + 7| + 4 < 1, and then |x + 7| + 4 ≥ 1.

   Show answer

   Both isolate to a negative number. |x + 7| < −3 has no solution, because an absolute value is never negative. |x + 7| ≥ −3 is true for all real numbers, (−∞, ∞).